Chap.4: EXAMPLE WITH SEVERAL AGGREGATESLet us go through a full, running sample problem which uses several aggregates. Below is the class diagram of the problem. In this simplified textual form, symbol A 0------x B represents an aggregate between A and B. This is the same as assigning many Bs to each A.
Each Department has several other Departments. In other words, Departments form a tree. Within this tree, each Department has a set of Employees and a set of Projects. The following program uses the class Aggregate from the Pattern Template Library. It creates a sample hierarchy of Departments, each with Employees and Projects, demonstrates how the data can be manipulated by removing one Employee and moving another Employee into a new position, and then prints the resulting data. Watch for the following details:
// ---------------------------------------------------------
#include <iostream.h>
#include <string.h>
#include <mgr.h> // always include
#include <aggregat.h>
class Department;
class Employee;
class Project;
class Department : public AggregateParent<Department,Department>,
public AggregateChild<Department,Department>,
public AggregateParent<Department,Employee>,
public AggregateParent<Department,Project>{
int mDeptNo;
void prtOffset(int i);
public:
Department(int n){mDeptNo=n;}
void prt(int offset); // print this and subordinate departments
Employee* find(char *name); // recursively, find this employee
};
class Employee : public AggregateChild<Department,Employee>{
char *mpName;
public:
Employee(char *n){mpName=new char[strlen(n)+1]; strcpy(mpName,n);}
void prt(void){cout << " " << mpName;}
int checkName(char *n){return !strcmp(mpName,n);}
};
class Project : public AggregateChild<Department,Project>{
char mProj;
public:
Project(char p){mProj=p;}
void prt(void){cout << " " << mProj;}
};
// ---------------------------------------------------
// The following lines declare the data organization.
// These objects do not have to be global, but making them
// global often contributes to the clarity of the code
// ---------------------------------------------------
Aggregate<Department,Department> departments;
Aggregate<Department,Employee> employees;
Aggregate<Department,Project> projects;
// ---------------------------------------------------
void Department::prtOffset(int i){
cout << "\n";
for(int k=0; k<i; k++) cout << " ";
}
void Department::prt(int offset){
Employee *pEmpl;
Project *pProj;
Department *pDept;
AggregateIterator<Department,Department> deptIt;
AggregateIterator<Department,Employee> emplIt;
AggregateIterator<Department,Project> projIt;
prtOffset(offset);
cout << "department= " << mDeptNo;
if(employees.head(this)){
prtOffset(offset);
cout << "employees:";
ITERATE(emplIt,this,pEmpl) pEmpl->prt();
}
if(projects.head(this)){
prtOffset(offset);
cout << "projects:";
ITERATE(projIt,this,pProj) pProj->prt();
}
cout << "\n";
// recursively, print all subordinated departments
ITERATE(deptIt,this,pDept) pDept->prt(offset+1);
}
Employee* Department::find(char *name){
Employee *pEmpl;
Department *pDept;
AggregateIterator<Department,Department> deptIt;
AggregateIterator<Department,Employee> emplIt;
// traverse the employees of this department
ITERATE(emplIt,this,pEmpl){
if(pEmpl->checkName(name))return pEmpl;
}
// recursively, search subordinate departments
ITERATE(deptIt,this,pDept){
pEmpl=pDept->find(name);
if(pEmpl)return pEmpl;
}
return NULL;
}
int main(void){
Department *pDept0,*pDept1,*pDept2;
Employee *pEmpl;
Project *pProj;
// create a small hierarchy of departments under pDept0
// To each department, add some employees and projects
pDept0=new Department(200);
pEmpl=new Employee("Head F."); employees.addTail(pDept0,pEmpl);
pDept1=new Department(220); departments.addTail(pDept0,pDept1);
pEmpl=new Employee("Mann H."); employees.addTail(pDept1,pEmpl);
pDept2=new Department(225); departments.addTail(pDept1,pDept2);
pEmpl=new Employee("Brown J."); employees.addTail(pDept2,pEmpl);
pEmpl=new Employee("Green D."); employees.addTail(pDept2,pEmpl);
pEmpl=new Employee("Black S."); employees.addTail(pDept2,pEmpl);
pProj=new Project('A'); projects.addTail(pDept2,pProj);
pDept2=new Department(228); departments.addTail(pDept1,pDept2);
pEmpl=new Employee("Little K."); employees.addTail(pDept2,pEmpl);
pEmpl=new Employee("Grossman A."); employees.addTail(pDept2,pEmpl);
pProj=new Project('B'); projects.addTail(pDept2,pProj);
pProj=new Project('C'); projects.addTail(pDept2,pProj);
pDept1=new Department(240); departments.addTail(pDept0,pDept1);
pEmpl=new Employee("Boss I."); employees.addTail(pDept1,pEmpl);
pDept2=new Department(243); departments.addTail(pDept1,pDept2);
pEmpl=new Employee("Zeller K."); employees.addTail(pDept2,pEmpl);
pEmpl=new Employee("Chapeau A."); employees.addTail(pDept2,pEmpl);
pEmpl=new Employee("Krpec P."); employees.addTail(pDept2,pEmpl);
pProj=new Project('D'); projects.addTail(pDept2,pProj);
pProj=new Project('E'); projects.addTail(pDept2,pProj);
pProj=new Project('F'); projects.addTail(pDept2,pProj);
// Boss I. leaves the company
pEmpl=pDept0->find("Boss I.");
employees.remove(pEmpl);
delete pEmpl;
// Chapeau A. is promoted as head of department 240
pEmpl=pDept2->find("Chapeau A.");
employees.remove(pEmpl);
employees.addTail(pDept1,pEmpl);
pDept0->prt(0); // print the entire organization
return(0);
}
/* ---------------------- results -----------------------
department= 200 employees: Head F. department= 220
employees: Mann H.
department= 225
employees: Brown J. Green D. Black S.
projects: A
department= 228
employees: Little K. Grossman A.
projects: B C
department= 240
employees: Chapeau A.
department= 243
employees: Zeller K. Krpec P.
projects: D E F
----------------------------------------------------*/
For your convenience, this program is stored in the documentation directory as PTL\DOC\TEST4.CPP, and the results are in file PTL\DOC\OUT4. |
